Question

Find the values of P such that rank of A=3 where
A=1 1 -1 0
4 4 -3 1
P 2 2 2
9 9 P 3​

Answer 1

Given:

             The rank of the Matrix = 3

To Find:

             The values of P

Solution:

The given matrix is,

                                $A =\left[\begin{array}{ccc}1&1&-1~~~0\\4&5&-3~~~1\\P&2&2~~~~~2\\9&9&P~~~~3\end{array}\right]$

Since the rank of a square matrix is negative of the determinant of that matrix.

So, the determinant of this matrix is,

                                $|A| = -3$

Now, For Finding the values of P,

We have to calculate the determinant.

$|A| = 1\left[\begin{array}{ccc}4&-3&1\\2&2&2\\9&P&3\end{array}\right] -1\left[\begin{array}{ccc}4&-3&1\\P&2&2\\9&P&3\end{array}\right]-1\left[\begin{array}{ccc}4&4&1\\P&2&2\\9&9&3\end{array}\right]-0$

   $= [4(6-2P)+3(6-18)+1(2P-18)]-[4(6-2P)+3(3P-18)+1(P^2-18)]-[4(6-18)-4(3P-18)+1(9P-18)]$

   $= [4(6-2P)-36+(2P-18)]-[4(6-2P)+3(3P-18)+(P^2-18)]-[-48-4(3P-18)+1(9P-18)]$

By Factorising the above equation,

                            $-3 = (P+6)(P-2)$

By solving the above equation, we get,

                                $P = -9,-1$

So, the values of P are -9 and -1